# Using the Schrodinger eqn in finding the momentum operator

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• Hamiltonian

#### Hamiltonian

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how can we use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?
I have read that the Schrodinger equation has no formal derivation we are simply applying the Hamiltonian operator on the wave function
$$\hat H = i\hbar \frac{\partial}{\partial t} = \hat T + \hat V$$
here we substitute $$\hat T = \frac{\hat p^2}{2m}$$ where $$\hat p = -i \hbar \frac{\partial}{\partial x}$$
but when we derive the equation for ##\hat p## we actually substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## from the Schrodinger equation.

$$< p> = m\frac{d<x>}{dt} = m\int_{-\infty}^{+\infty} x\frac{\partial (\psi^*\psi)}{\partial t}$$
$$<p> = m\int_{-\infty}^{+\infty} x[\psi^*\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}] dx$$
here we substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## as
$$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} -\frac{i}{\hbar} V\psi$$
$$\frac{\partial \psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} +\frac{i}{\hbar} V\psi^*$$
after some simplification we end up with
$$<p> = \int_{-\infty}^{+\infty} \psi^* (-i\hbar \frac{\partial}{\partial x})\psi dx$$
and then finally we get $$\hat p = -i\hbar \frac{\partial }{\partial x}$$
so I don't understand how we can use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?
this video does the derivation for the momentum operator

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What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where ##\hat p = -i\hbar \frac{\partial}{\partial x}##. And that justifies the original definition of ##\hat p##.

• vanhees71
What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where ##\hat p = -i\hbar \frac{\partial}{\partial x}##. And that justifies the original definition of ##\hat p##.
how exactly do we find ##\hat p## without using the Schrodinger equation? by finding ##\hat p## I mean how do we arrive at ##\hat p = -i\hbar \frac{\partial}{\partial x}##. I thought the only way of arriving at this would be by using ##m\frac{d<x>}{dt} = <p>## but when we use this approach we need to use the schrodinger equation but the KE energy term in the Hamiltonian is already using ##\hat p = -i\hbar \frac{\partial}{\partial x}##

• vanhees71
how exactly do we find ##\hat p## without using the Schrodinger equation? by finding ##\hat p## I mean how do we arrive at ##\hat p = -i\hbar \frac{\partial}{\partial x}##. I thought the only way of arriving at this would be by using ##m\frac{d<x>}{dt} = <p>## but when we use this approach we need to use the schrodinger equation but the KE energy term in the Hamiltonian is already using ##\hat p = -i\hbar \frac{\partial}{\partial x}##
There are ways to motivate the definition of momentum. For example:

https://en.wikipedia.org/wiki/Momentum_operator#Origin_from_De_Broglie_plane_waves

And also on that page momentum as the generator of spatial translations (this is done in Sakurai's book).

• vanhees71 and Hamiltonian